Download Analysis I - IV by Hilgert J. PDF

By Hilgert J.

Der vorliegende textual content ist eine vorlaufige Ausarbeitung meiner Vorlesungen research I-IV (Wintersemester 2004/2005 { Sommersemester 2006) an der Universitat Paderborn.

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6 : Bestimme die gr¨ oßte untere und die kleinste obere Schranke der folgenden Mengen: (i) X := { n1 | n ∈ N}. (ii) X := {x ∈ R | x < 1}. 7 : Es sei (Z, +, ·, P ) ein geordneter K¨ orper. Man zeige: Das Supremum der Menge aller unteren Schranken von P ist 0. 8 : Genau die Zahlen aus R \ Q heißen irrationale Zahlen. Zeige, daß zwischen zwei reellen Zahlen immer eine rationale und eine irrationale Zahl liegen. Man sagt dann auch, daß Q und R \ Q dicht in R liegen. B. hat {x ∈ R | x ≤ 2} in Q kein Supremum.

Dann gilt (vgl. 5) 1 t |g(t) − y| = |f 1 t > x0 f¨ ur 0 < t < δ und − y| < f¨ ur t ∈ ]0, δ[ ∩N . Dies bedeutet aber, daß limt→0+ g(t) = y. Umgekehrt, wenn limt→0+ g(t) = y, dann gilt (∗∗) f¨ ur t ∈ ]0, δ[ ∩N . Mit x0 = dies (∗) f¨ ur x ∈ ]x0 , ∞[ ∩M , also limx→∞ f (x) = y. 9 beweist man: Sei M ⊆ R nach oben unbeschr¨ankt und f : M → R eine Funktion. Setze N := { x1 | x ∈ M \ {0}} und g : N → R, t → f ( 1t ). Dann gilt ] − δ, 0[ ∩N = ∅ f¨ ur jedes δ > 0 und lim f (x) = y x→−∞ ⇔ lim g(t) = y. 11 : x2 − 2x = lim x→∞ 2x2 − 3x − 1 x→0+ lim 2 x2 1 x2 − − 3 x 2 x −1 1 − 2t 1 = 2 t→0+ 2 − 3t − t 2 = lim Im allgemeinen haben Funktionen keinen Grenzwert f¨ ur x → ±∞.

Dann finden wir y 1 x < f¨ ur alle x > x+ und f¨ ur alle x < x− . ✻ .. ... ... ...... ............... ................... ....... ... ... .. ✲x 40 KAPITEL 2. 5 : Die Menge {f ∈ F ]a, ∞[, R | ∃ lim f (x)} x→∞ ¨ ist ein Untervektorraum von F(]a, ∞[, R) (vgl. 9 : Sei M ⊆ R nach oben unbeschr¨ ankt und f : M → R eine Funktion. Setze N := { x1 | x ∈ M \ {0}} und g : N → R, t → f ( 1t ). Dann gilt ]0, δ[ ∩N = ∅ f¨ ur jedes δ > 0 und lim f (x) = y ⇔ x→∞ lim g(t) = y. 5 und den Definitionen.

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